<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<!--Converted with LaTeX2HTML 98.2 beta6 (August 14th, 1998)
original version by:  Nikos Drakos, CBLU, University of Leeds
* revised and updated by:  Marcus Hennecke, Ross Moore, Herb Swan
* with significant contributions from:
  Jens Lippmann, Marek Rouchal, Martin Wilck and others -->
<HTML>
<HEAD>
<TITLE>Further Details:
Error Bounds for Linear Equality Constrained Least Squares Problems</TITLE>
<META NAME="description" CONTENT="Further Details:
Error Bounds for Linear Equality Constrained Least Squares Problems">
<META NAME="keywords" CONTENT="lug_l2h">
<META NAME="resource-type" CONTENT="document">
<META NAME="distribution" CONTENT="global">
<META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1">
<LINK REL="STYLESHEET" HREF="lug_l2h.css">
<LINK REL="previous" HREF="node85.html">
<LINK REL="up" HREF="node85.html">
<LINK REL="next" HREF="node87.html">
</HEAD>
<BODY >
<!--Navigation Panel-->
<A NAME="tex2html5398"
 HREF="node87.html">
<IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next"
 SRC="next_motif.gif"></A> 
<A NAME="tex2html5392"
 HREF="node85.html">
<IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up"
 SRC="up_motif.gif"></A> 
<A NAME="tex2html5388"
 HREF="node85.html">
<IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous"
 SRC="previous_motif.gif"></A> 
<A NAME="tex2html5394"
 HREF="node1.html">
<IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents"
 SRC="contents_motif.gif"></A> 
<A NAME="tex2html5396"
 HREF="node152.html">
<IMG WIDTH="43" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="index"
 SRC="index_motif.gif"></A> 
<BR>
<B> Next:</B> <A NAME="tex2html5399"
 HREF="node87.html">General Linear Model Problem</A>
<B> Up:</B> <A NAME="tex2html5393"
 HREF="node85.html">Linear Equality Constrained Least</A>
<B> Previous:</B> <A NAME="tex2html5389"
 HREF="node85.html">Linear Equality Constrained Least</A>
 &nbsp <B>  <A NAME="tex2html5395"
 HREF="node1.html">Contents</A></B> 
 &nbsp <B>  <A NAME="tex2html5397"
 HREF="node152.html">Index</A></B> 
<BR>
<BR>
<!--End of Navigation Panel-->

<H3><A NAME="SECTION03461100000000000000"></A>
<A NAME="sec_lseglm_lsedetails"></A>
<BR>
Further Details:
Error Bounds for Linear Equality Constrained Least Squares Problems
</H3>

<P>
In this subsection, we will summarize the available error bound.
The reader may also refer to [<A
 HREF="node151.html#lawn31">2</A>,<A
 HREF="node151.html#baifahey97">13</A>,<A
 HREF="node151.html#coxhigham">18</A>,<A
 HREF="node151.html#elden">50</A>] for
further details.  
<BR>

<P>
Let <IMG
 WIDTH="14" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
 SRC="img467.gif"
 ALT="$\widehat {x}$">
be the solution computed by the driver xGGLSE (see subsection
<A HREF="node84.html#sec_lseglm_drivers">4.6</A>).  It is normwise
stable in a mixed forward/backward sense [<A
 HREF="node151.html#coxhigham">18</A>,<A
 HREF="node151.html#baifahey97">13</A>].
Specifically, 
<!-- MATH
 $\widehat {x} = \bar {x} + \Delta \bar {x}$
 -->
<IMG
 WIDTH="93" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img475.gif"
 ALT="$\widehat {x} = \bar {x} + \Delta \bar {x}$">,
where <IMG
 WIDTH="14" HEIGHT="16" ALIGN="BOTTOM" BORDER="0"
 SRC="img476.gif"
 ALT="$\bar {x}$">
solves

<!-- MATH
 $\min\{\| b + \Delta b - (A + \Delta A)x \|_2: \; (B + \Delta B)x = d \}$
 -->
<IMG
 WIDTH="373" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
 SRC="img477.gif"
 ALT="$\min\{\Vert b + \Delta b - (A + \Delta A)x \Vert _2: \; (B + \Delta B)x = d \} $">,
and
<DIV ALIGN="CENTER">
<IMG
 WIDTH="234" HEIGHT="69" ALIGN="MIDDLE" BORDER="0"
 SRC="img478.gif"
 ALT="$\textstyle \parbox{2in}{
\begin{eqnarray*}
\Vert \Delta \bar {x} \Vert _2 &amp; \...
... \Delta A \Vert _F &amp; \leq &amp; q(m,n,p)\epsilon\Vert A\Vert _F,
\end{eqnarray*} }$">&nbsp;&nbsp;&nbsp;&nbsp;
<IMG
 WIDTH="234" HEIGHT="69" ALIGN="MIDDLE" BORDER="0"
 SRC="img479.gif"
 ALT="$\textstyle \parbox{2in}{
\begin{eqnarray*}
\Vert \Delta b \Vert _2 &amp; \leq &amp; q...
...t \Delta B \Vert _F &amp; \leq &amp; q(m,n,p)\epsilon\Vert B\Vert _F,
\end{eqnarray*}}$">
</DIV>
<B><I>q</I>(<I>m</I>,<I>n</I>,<I>p</I>)</B> is a modestly growing function of <B><I>m</I></B>, <B><I>n</I></B>, and <B><I>p</I></B>.
We take <B><I>q</I>(<I>m</I>,<I>n</I>,<I>p</I>) = 1</B> in the code fragment above.
Let <IMG
 WIDTH="27" HEIGHT="18" ALIGN="BOTTOM" BORDER="0"
 SRC="img480.gif"
 ALT="$X^{\dagger}$">
denote the Moore-Penrose pseudo-inverse of <B><I>X</I></B>.
Let 
<!-- MATH
 $\kappa_B(A) =  \| A \|_F  \| (AP)^\dagger \|_2$
 -->
<IMG
 WIDTH="190" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
 SRC="img481.gif"
 ALT="$\kappa_B(A) = \Vert A \Vert _F \Vert (AP)^\dagger \Vert _2 $">( = <TT>CNDAB</TT> above) and
    
<!-- MATH
 $\kappa_A(B) =  \| B \|_F  \| B^\dagger_A \|_2$
 -->
<IMG
 WIDTH="168" HEIGHT="42" ALIGN="MIDDLE" BORDER="0"
 SRC="img482.gif"
 ALT="$\kappa_A(B) = \Vert B \Vert _F \Vert B^\dagger_A \Vert _2 $">( = <TT>CNDBA</TT> above)
where 
<!-- MATH
 $P = I - B^\dagger B$
 -->
<IMG
 WIDTH="107" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
 SRC="img483.gif"
 ALT="$P = I - B^\dagger B$">
and 
<!-- MATH
 $B^\dagger_A = (I - (AP)^\dagger A)B^\dagger$
 -->
<IMG
 WIDTH="177" HEIGHT="42" ALIGN="MIDDLE" BORDER="0"
 SRC="img484.gif"
 ALT="$B^\dagger_A = (I - (AP)^\dagger A)B^\dagger$">.
When 
<!-- MATH
 $q(m,n,p)\epsilon$
 -->
<IMG
 WIDTH="84" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
 SRC="img485.gif"
 ALT="$q(m,n,p)\epsilon$">
is small, the error <IMG
 WIDTH="46" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
 SRC="img415.gif"
 ALT="$x-\hat{x}$">
is bounded by
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
\frac{ \| x-\hat{x} \| }{ \| x \| } \def \theguybelow{\sim}
\def \verticalposition{\lower 2.5pt}
\def \spacingwithinsymbol{\baselineskip0pt.2pt}
\mathrel{\mathpalette\verticalposition\vbox{\spacingwithinsymbol
\everycr={}\tabskip0pt
\halign{$;\hfil##\hfil$\crcr S\crcr
\theguybelow\crcr}}PMlt;}q(m,n,p)\epsilon\left\{
      \kappa_A(B) +
      \kappa_B(A)\left(\frac{ \| b \|_2 }{ \| A \|_F  \| x \|_2 } + 1 \right) +
      \kappa^2_B(A)\left( \frac{ \| B \|_F }{ \| A \|_F } \| AB^\dagger_A \|_2  +
                     1\right)\frac{ \| r \|_2 }{ \| A \|_F  \| x \|_2 } \right\}.
\end{displaymath}
 -->


<IMG
 WIDTH="777" HEIGHT="72" BORDER="0"
 SRC="img486.gif"
 ALT="\begin{displaymath}
% latex2html id marker 15239\frac{ \Vert x-\hat{x} \Vert ...
...rt r \Vert _2 }{ \Vert A \Vert _F \Vert x \Vert _2 } \right\}.
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>

<P>
<BR>
<BR>

<P>
When <B><I>B</I> = 0</B> and <B><I>d</I> = 0</B>, we essentially recover error bounds for the
linear least squares (LS) problem:
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
\frac{ \| x-\widehat {x} \|_2 }{ \| x \|_2 } \leq q(m,n)\epsilon
      \left\{\kappa(A)\left(
          \frac{ \| b \|_2 }{ \| A \|_F  \| x \|_2 } + 1 +
          \kappa(A)\frac{ \| r \|_2 }{ \| A \|_F  \| x \|_2 } \right) \right\},
\end{displaymath}
 -->


<IMG
 WIDTH="500" HEIGHT="48" BORDER="0"
 SRC="img487.gif"
 ALT="\begin{displaymath}
\frac{ \Vert x-\widehat {x} \Vert _2 }{ \Vert x \Vert _2 } ...
...rt _2 }{ \Vert A \Vert _F \Vert x \Vert _2 } \right) \right\},
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where 
<!-- MATH
 $\kappa(A) =  \| A \|_F  \| A^\dagger \|_2$
 -->
<IMG
 WIDTH="151" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
 SRC="img488.gif"
 ALT="$\kappa(A) = \Vert A \Vert _F \Vert A^\dagger \Vert _2 $">.
Note that
the error in the standard least squares problem provided in section 4.5.1 is
<BR><P></P>
<DIV ALIGN="CENTER">

<!-- MATH
 \begin{displaymath}
\frac{ \| x-\widehat {x} \|_2 }{ \| x \|_2 } \def \theguybelow{\sim}
\def \verticalposition{\lower 2.5pt}
\def \spacingwithinsymbol{\baselineskip0pt.2pt}
\mathrel{\mathpalette\verticalposition\vbox{\spacingwithinsymbol
\everycr={}\tabskip0pt
\halign{$;\hfil##\hfil$\crcr S\crcr
\theguybelow\crcr}}PMlt;}p(n)\epsilon
      \left\{\frac{2\kappa(A)}{\cos(\theta)} + \tan(\theta)\kappa^2(A)\right\}
      =  p(n)\epsilon \left\{\frac{2\kappa(A) \| b \|_2 }{ \| Ax \|_2 } +
                  \kappa^2(A)\frac{ \| r \|_2 }{ \| Ax \|_2 }\right\}\\
\end{displaymath}
 -->


<IMG
 WIDTH="555" HEIGHT="112" BORDER="0"
 SRC="img489.gif"
 ALT="\begin{displaymath}
% latex2html id marker 13969\frac{ \Vert x-\widehat {x} \...
...^2(A)\frac{ \Vert r \Vert _2 }{ \Vert Ax \Vert _2 }\right\}\\
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
since 
<!-- MATH
 $\sin(\theta) = \frac{ \| A\widehat {x}-b \|_2 }{ \| b \|_2 }$
 -->
<IMG
 WIDTH="129" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"
 SRC="img490.gif"
 ALT="$\sin(\theta) = \frac{ \Vert A\widehat {x}-b \Vert _2 }{ \Vert b \Vert _2 }$">.
If one assumes
that 
<!-- MATH
 $q(m,n) = p(n) = 1$
 -->
<B><I>q</I>(<I>m</I>,<I>n</I>) = <I>p</I>(<I>n</I>) = 1</B>, then the bounds are essentially the same. 
<BR>

<P>
<HR>
<!--Navigation Panel-->
<A NAME="tex2html5398"
 HREF="node87.html">
<IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next"
 SRC="next_motif.gif"></A> 
<A NAME="tex2html5392"
 HREF="node85.html">
<IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up"
 SRC="up_motif.gif"></A> 
<A NAME="tex2html5388"
 HREF="node85.html">
<IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous"
 SRC="previous_motif.gif"></A> 
<A NAME="tex2html5394"
 HREF="node1.html">
<IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents"
 SRC="contents_motif.gif"></A> 
<A NAME="tex2html5396"
 HREF="node152.html">
<IMG WIDTH="43" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="index"
 SRC="index_motif.gif"></A> 
<BR>
<B> Next:</B> <A NAME="tex2html5399"
 HREF="node87.html">General Linear Model Problem</A>
<B> Up:</B> <A NAME="tex2html5393"
 HREF="node85.html">Linear Equality Constrained Least</A>
<B> Previous:</B> <A NAME="tex2html5389"
 HREF="node85.html">Linear Equality Constrained Least</A>
 &nbsp <B>  <A NAME="tex2html5395"
 HREF="node1.html">Contents</A></B> 
 &nbsp <B>  <A NAME="tex2html5397"
 HREF="node152.html">Index</A></B> 
<!--End of Navigation Panel-->
<ADDRESS>
<I>Susan Blackford</I>
<BR><I>1999-10-01</I>
</ADDRESS>
</BODY>
</HTML>
